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ikadub [295]
3 years ago
15

Comment Both propane and benzene are hydrocarbons. As a rule,

Chemistry
1 answer:
kozerog [31]3 years ago
5 0

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

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1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm


Ideal gas equation: pV = nRT => n = pV / RT


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V = 300 cm^3 = 0.300 liter

T = 298 K

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=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol


2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type


X (+) + O2 (g) ---> X2O          or   


2 X(2+) + O2(g) ----> X2O2 = 2XO     or


4X(3+) + 3O2(g) ---> 2X2O3


 
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)



So, lets probe those 3 cases.


3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

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So, now probe the case 2.



4) Case 2:


2moles X metal / 1 mol O2(g) = x / 0.01226 mol


=> x = 2 * 0.01226 = 0.02452 mol


And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol


That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.


4) Case 3


4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635 


atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol


That does not correspond to any metal.


Conclusion: the identity of the metallic element could be titanium.
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